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About geodesic coordinates:

Let S be regular surface. $p\in S$

$\gamma$ be unit speed geodesic on $S$ with parameter $v$ and $\gamma (0)=p$

$\tilde \gamma^v$ be unit speed geodesic s.t. $\tilde \gamma^v(0)=\gamma(v)$

$\tilde \gamma^v$ is perpendicular to $\gamma$ at $\gamma (v)$.

Define $\sigma : U\to S$ $\sigma (u,v)=\tilde \gamma^v (u)$

In this coordinate system first fundamental form is $ds^2=du^2+G(u,v)dv^2$ Where $G(0,v)=1 $ and $G_u(0,v)=0$

Thank you for helping.

  • $\begingroup$ @ B11b: hey, this is the same as your other question, math.stackexchange.com/questions/726386/…, on which I'm currently working! Indeed, writing up my answer! I know you want an answer pronto, but I dare to suggest the duping a question is not the best way to reach that goal! So wassup? $\endgroup$ – Robert Lewis Mar 25 '14 at 20:09
  • 1
    $\begingroup$ @ B11b: OK, you deleted the other version, good idea! I'll transfer the write-up I'm working on to this version of the question, no problem. But I would like to ask you to refrain from posting duplicates in such manner, and I think many here would agree. OK, enough said. $\endgroup$ – Robert Lewis Mar 25 '14 at 20:29
  • $\begingroup$ Okay i didnt know. Thank you :) @RobertLewis $\endgroup$ – user315 Mar 25 '14 at 20:50
  • $\begingroup$ @ B11b: OK, 'sall good. I'm typing something up on it right now but I may have to leave for work before I finish, so perhaps tonight I'll be able to post something. Will keep you "posted", as it were! Regards. ;-) $\endgroup$ – Robert Lewis Mar 25 '14 at 20:57
  • $\begingroup$ I dont need an answer in detail. Thanks:) @RobertLewis $\endgroup$ – user315 Mar 25 '14 at 21:44

Assuming that the procedure described in the body of the question actually does define a coordinate patch (I haven't checked, but it seems quite plausible so I'll accept its validity), then, as far as finding the geodesic equations is concerned, we are reduced to the simple question of finding them for a metric of the form

$\hat g = ds^2 = du^2 + G(u, v)dv^2, \tag{1}$

where $G(u, v)$ meets the presented conditions:

$G(0, v) = 1 \; \text{and} \; G_u(0, v) = 0. \tag{2}$

These equations are derived in my answer to this one of B11b's other questions; we have

$\ddot u + \dot u^2 \Gamma_{11}^1 + 2\dot u \dot v \Gamma_{12}^1 + \dot v^2 \Gamma_{22}^1 = 0 \tag{3}$


$\ddot v + \dot u^2 \Gamma_{11}^2 + 2\dot u \dot v \Gamma_{12}^2 + \dot v^2 \Gamma_{22}^2 = 0, \tag{4}$

where the $\Gamma_{\mu \nu}^\alpha$, $\alpha, \mu, \nu = 1, 2$, are the Christoffel symbols for the covariant derivative associated to the metric tensor field $ds^2$ of (1). The $\Gamma_{\mu \nu}^\alpha$ may be simply had from the components of $\hat g = ds^2$ in via well-knonw formulas which are given below; before proceeding in that direction, however, I pause for an adjustment of notation. In (3) and (4), $u$ is the $1$-coordinate and $v$ is the $2$-coordinate in the sense that $u$ corresponds to the index "$1$" and $v$ to index "$2$" on the Christoffel symbols. I now define/set $x^1 = u$ and $x^2 = v$, and will use the $x^i$, $1 = 1, 2$, in place of $u, v$ for the following arguments/calulations, since it is notationally much more convenient to deal with variables indexed/denoted in a consistent manner throughout; at the end, we can write any results in terms of $u, v$ if so desired.

The coefficients $\Gamma_{\mu \nu}^\alpha$ occurring in (3) and (4) are related to the components $g_{\mu \nu}$ of the metric $\hat g$ according to

$\Gamma_{\mu \nu}^\alpha = \dfrac{1}{2}g^{\alpha \rho}(g_{\rho \mu, \nu} + g_{\rho \nu, \mu} - g_{\mu \nu, \rho}), \tag{5}$

wherein the coefficients $g^{\alpha \rho}$ are the components of the inverse of the matrix representing the tensor field $\hat g$ in $x^1, x^2$ coordinates (see for example this widipedia entry). In the present case, since $\hat g$ takes such a simple form

$ds^2 = \hat g = [g_{\mu \nu}] = \begin{bmatrix} 1 & 0 \\ 0 & G(u, v) \end{bmatrix}, \tag{6}$

we have

$[g^{\mu \nu}] = \begin{bmatrix} 1 & 0 \\ 0 & G^{-1}(u, v) \end{bmatrix}, \tag{7}$

and since $\hat g$ has only one non-constant component $g_{22} = G(u, v)$, calculation of the $\Gamma_{\mu \nu}^\alpha$ becomes a relatively simple matter; indeed, since $g^{\alpha \beta} = 0$ for $\alpha \ne \beta$, and since $g_{\mu \nu, \sigma} = 0$ unless $\mu = \nu = 2$, we see from (5) that

$\Gamma_{22}^1 = \dfrac{1}{2}g^{11}(-g_{22, 1}) = -\dfrac{1}{2}G_{, x_1}; \tag{8}$

$\Gamma_{\mu \nu}^1 = 0 \tag{9}$

if $\mu = 1$ or $\nu = 1$,


$\Gamma_{11}^2 = 0; \tag{10}$

$\Gamma_{12}^2 = \Gamma_{21}^2 = \dfrac{1}{2}g^{22}g_{22, 1} = \dfrac{1}{2}G_{-1}G_{, x_1}; \tag{11}$

$\Gamma_{22}^2 = \dfrac{1}{2}g^{22}g_{22, 2} = \dfrac{1}{2}G^{-1}G_{, x_2}, \tag{12}$

wherein $G_{, x_1} = \partial G / \partial x_1$ and so forth; all the other $\Gamma_{\mu \nu}^\alpha$ vanish. Apparently, then, the geodesic equations (3)-(4) take the particularly simple form

$\ddot x^1 - \dfrac{1}{2}G_{, x_1}\dot x_2^2 = 0, \tag{13}$

$\ddot x^2 + G^{-1}G_{, x_1}\dot x_1 \dot x_2 + \dfrac{1}{2}G^{-1}G_{, x_2}\dot x_2^2 = 0, \tag{14}$

or, if you like, converted back into the $u, v$ notation:

$\ddot u - \dfrac{1}{2}G_{, u}\dot v^2 = 0, \tag{15}$

$\ddot v + G^{-1}G_{, u} \dot u \dot v + \dfrac{1}{2}G^{-1}G_{, v}\dot v^2 = 0; \tag{16}$

(15) and (16) are indeed relatively simple as geodesic equations go.

As for the length $l(\gamma, t_0, t_1)$ of a curve $\gamma (t)$ twixt $\gamma(t_0)$ and $\gamma (t_1)$, i.e., for $\gamma:[t_0, t_1] \to S$, it is as usual given by the integral of the magnitude of the tangent vector $\gamma'(t)$, $\vert \gamma' \vert = \sqrt{\hat g(\dot \gamma, \dot \gamma)}$:

$l(\gamma, t_0, t_1) = \int_{t_0}^{t_1} \sqrt{\hat g(\dot \gamma, \dot \gamma)} dt; \tag{17}$

in the $u$-$v$ coordinate system, with the metric given by (1), this expression also takes a particularly simple form; we have, writing the coordinate expression of $\gamma(t) = (\gamma^u(t), \gamma^v(t))$, so that $\dot \gamma(t) = \dot \gamma^u(t) (\partial / \partial u) + \dot \gamma^v(t) (\partial / \partial v) $,

$\hat g(\dot \gamma(t), \dot \gamma(t)) = (\dot \gamma^u(t))^2 + G(\gamma^u(t), \gamma^v(t))(\dot \gamma^v(t))^2, \tag{18}$


$l(\gamma, t_0, t_1) = \int_{t_0}^{t_1} \sqrt{(\dot \gamma^u(t))^2 + G(\gamma^u(t), \gamma^v(t))(\dot \gamma^v(t))^2} dt, \tag{19}$

itself a relatively simple expression as such things go.

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

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    $\begingroup$ @ B11b: and you are most welcome, sir! Always glad to help out! And thanks for the "acceptance"! $\endgroup$ – Robert Lewis Mar 26 '14 at 22:06
  • $\begingroup$ OK, no more edits! $\endgroup$ – Robert Lewis Mar 27 '14 at 6:14

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